Calculate the area of the oval along the perimeter. Oval

Calculating the length/perimeter of an ellipse is not at all a trivial task, as one might think.

But the same simple approach is completely unsuitable for an ellipse.

In exact expression, the perimeter of an ellipse can only be expressed through this formula:

Ellipse eccentricity

Semimajor axis of the ellipse

In everyday life, of course, approximate formulas are used, which we will talk about.

One of them looks like this

The formula gives twice as accurate data

And an even more accurate perimeter of the ellipse gives the expression

But, no matter what the formulas are, they still only approximately give the perimeter of the ellipse.

We, using an exact formula through the elliptic integral, obtain independence from such restrictions, and obtain absolute accuracy for any value of the ellipse.

Solving Examples

The ellipse is given by the equation

Find its perimeter

Let's enter the known parameters a=2 and b=5 and get the result

Why can only semi-axes values be entered in the source data? According to other parameters, what doesn’t count?

I'll explain.

The calculators on this site, including this one, are not intended to replace your brain. They only simplify routine operations, or those operations where it is possible to make a mistake. And that's all.

Circumference is a closed plane curve, all points of which are equidistant from a given point (the center of the circle). The distance from any point of the circle \(P\left((x,y) \right)\) to its center is called radius. The center of the circle and the circle itself lie in the same plane. Equation of a circle of radius \(R\) with center at the origin ( canonical equation of a circle ) has the form
\((x^2) + (y^2) = (R^2)\).

Equation of a circle radius \(R\) with center at an arbitrary point \(A\left((a,b) \right)\) is written as
\((\left((x - a) \right)^2) + (\left((y - b) \right)^2) = (R^2)\).

Equation of a circle passing through three points , written in the form: \(\left| (\begin(array)(*(20)(c)) ((x^2) + (y^2)) & x & y & 1\\ (x_1^2 + y_1^2) & ((x_1)) & ((y_1)) & 1\\ (x_2^2 + y_2^2) & ((x_2)) & ((y_2)) & 1\\ (x_3^2 + y_3^2) & ((x_3)) & ((y_3)) & 1 \end(array)) \right| = 0.\\\)
Here \(A\left(((x_1),(y_1)) \right)\), \(B\left(((x_2),(y_2)) \right)\), \(C\left(( (x_3),(y_3)) \right)\) are three points lying on the circle.

Equation of a circle in parametric form
\(\left\( \begin(aligned) x &= R \cos t \\ y &= R\sin t \end(aligned) \right., \;\;0 \le t \le 2\pi\ ),
where \(x\), \(y\) are the coordinates of the points of the circle, \(R\) is the radius of the circle, \(t\) is the parameter.

General equation of a circle
\(A(x^2) + A(y^2) + Dx + Ey + F = 0\)
subject to \(A \ne 0\), \(D^2 + E^2 > 4AF\).
The center of the circle is located at the point with coordinates \(\left((a,b) \right)\), where
\(a = - \large\frac(D)((2A))\normalsize,\;\;b = - \large\frac(E)((2A))\normalsize.\)
The radius of the circle is
\(R = \sqrt (\large\frac(((D^2) + (E^2) - 4AF))((2\left| A \right|))\normalsize) \)

Ellipse is a plane curve for each point of which the sum of the distances to two given points ( ellipse foci ) is constant. The distance between the foci is called focal length and is denoted by \(2c\). The middle of the segment connecting the foci is called the center of the ellipse . An ellipse has two axes of symmetry: the first or focal axis, passing through the foci, and the second axis perpendicular to it. The points of intersection of these axes with the ellipse are called peaks. The segment connecting the center of the ellipse with the vertex is called semi-axis of the ellipse . The semi-major axis is denoted by \(a\), the semi-minor axis by \(b\). An ellipse whose center is at the origin and whose semi-axes lie on coordinate lines is described by the following canonical equation :
\(\large\frac(((x^2)))(((a^2)))\normalsize + \large\frac(((y^2)))(((b^2)))\ normalsize = 1.\)

The sum of the distances from any point of the ellipse to its foci constant:
\((r_1) + (r_2) = 2a\),
where \((r_1)\), \((r_2)\) are the distances from an arbitrary point \(P\left((x,y) \right)\) to the foci \((F_1)\) and \(( F_2)\), \(a\) is the semimajor axis of the ellipse.

The relationship between the semi-axes of the ellipse and the focal length
\((a^2) = (b^2) + (c^2)\),
where \(a\) is the semi-major axis of the ellipse, \(b\) is the semi-minor axis, \(c\) is half the focal length.

Ellipse eccentricity
\(e = \large\frac(c)(a)\normalsize

Equations of ellipse directrixes
The directrix of an ellipse is a straight line perpendicular to its focal axis and intersecting it at a distance \(\large\frac(a)(e)\normalsize\) from the center. The ellipse has two directrixes located on opposite sides of the center. The directrix equations are written in the form
\(x = \pm \large\frac(a)(e)\normalsize = \pm \large\frac(((a^2)))(c)\normalsize.\)

Equation of an ellipse in parametric form
\(\left\( \begin(aligned) x &= a\cos t \\ y &= b\sin t \end(aligned) \right., \;\;0 \le t \le 2\pi\ ),
where \(a\), \(b\) are the semi-axes of the ellipse, \(t\) is the parameter.

General equation of ellipse
\(A(x^2) + Bxy + C(y^2) + Dx + Ey + F = 0\),
where \((B^2) - 4AC

General equation of an ellipse whose semi-axes are parallel to the coordinate axes
\(A(x^2) + C(y^2) + Dx + Ey + F = 0\),
where \(AC > 0\).

Ellipse perimeter
\(L = 4aE\left(e \right)\),
where \(a\) is the semimajor axis of the ellipse, \(e\) is the eccentricity, \(E\) is complete elliptic integral of the second kind.

Approximate formulas for the perimeter of an ellipse
\(L \approx \pi \left[ (\large\frac(3)(2)\normalsize\left((a + b) \right) - \sqrt (ab) ) \right],\;\;L \approx \pi \sqrt (2\left(((a^2) + (b^2)) \right)),\)
where \(a\), \(b\) are the semi-axes of the ellipse.

Area of the ellipse
\(S = \pi ab\)

In astronomy, when considering the movement of cosmic bodies in orbits, the concept of “ellipse” is often used, since their trajectories are characterized by precisely this curve. In the article we will consider the question of what the marked figure represents, and also give the formula for the length of the ellipse.

What is an ellipse?

According to the mathematical definition, an ellipse is a closed curve for which the sum of the distances from any of its points to two other specific points lying on the main axis, called foci, is a constant value. Below is a figure that explains this definition.

In the figure, the sum of the distances PF" and PF is equal to 2 * a, that is, PF" + PF = 2 * a, where F" and F are the foci of the ellipse, "a" is the length of its semi-major axis. The segment BB" is called the semi-minor axis, and distance CB = CB" = b - length of the semi-minor axis. Here point C determines the center of the figure.

The picture above also shows a simple rope and two nails method that is widely used for drawing elliptic curves. Another way to obtain this figure is to perform it at any angle to its axis, which is not equal to 90 o.

If the ellipse is rotated along one of its two axes, then it forms a three-dimensional figure, which is called a spheroid.

Formula for the circumference of an ellipse

Although the figure in question is quite simple, the length of its circumference can be accurately determined by calculating the so-called elliptic integrals of the second kind. However, the self-taught Hindu mathematician Ramanujan, at the beginning of the 20th century, proposed a fairly simple formula for the length of an ellipse, which approaches the result of the marked integrals from below. That is, the value of the value in question calculated from it will be slightly less than the actual length. This formula looks like: P ≈ pi *, where pi = 3.14 is the number pi.

For example, let the lengths of the two semi-axes of the ellipse be equal to a = 10 cm and b = 8 cm, then its length P = 56.7 cm.

Everyone can check that if a = b = R, that is, an ordinary circle is considered, then Ramanujan’s formula reduces to the form P = 2 * pi * R.

Note that in school textbooks another formula is often given: P = pi * (a + b). It is simpler, but also less accurate. So, if we apply it to the case considered, we obtain the value P = 56.5 cm.

What is an ellipse?

Formula for the circumference of an ellipse

For example, let the lengths of the two semi-axes of the ellipse be equal to a = 10 cm and b = 8 cm, then its length P = 56.7 cm.

Everyone can check that if a = b = R, that is, an ordinary circle is considered, then Ramanujan’s formula reduces to the form P = 2 * pi * R.

Note that in school textbooks another formula is often given: P = pi * (a + b). It is simpler, but also less accurate. So, if we apply it to the case considered, we obtain the value P = 56.5 cm.

Oval is a closed box curve that has two axes of symmetry and consists of two support circles of the same diameter, internally conjugate by arcs (Fig. 13.45). An oval is characterized by three parameters: length, width and radius of the oval. Sometimes only the length and width of the oval are specified, without defining its radii, then the problem of constructing an oval has a large variety of solutions (see Fig. 13.45, a... d).

Methods for constructing ovals based on two identical reference circles that touch (Fig. 13.46, a), intersect (Fig. 13.46, b) or do not intersect (Fig. 13.46, c) are also used. In this case, two parameters are actually specified: the length of the oval and one of its radii. This problem has many solutions. It's obvious that R > OA has no upper bound. In particular R = O 1 O 2(see Fig. 13.46.a, and Fig. 13.46.c), and the centers O 3 And O 4 are determined as the points of intersection of the base circles (see Fig. 13.46, b). According to the general point theory, mates are determined on a straight line connecting the centers of arcs of osculating circles.

Constructing an oval with touching support circles(the problem has many solutions) ( rice. 3.44). From the centers of the reference circles ABOUT And 0 1 with a radius equal, for example, to the distance between their centers, arcs of circles are drawn until they intersect at points ABOUT 2 and O 3.

Figure 3.44

If from points ABOUT 2 and O 3 draw straight lines through the centers ABOUT And O 1, then at the intersection with the support circles we obtain the connecting points WITH, C 1, D And D 1. From points ABOUT 2 and O 3 as from centers of radius R 2 draw arcs of conjugation.

Constructing an oval with intersecting reference circles(the problem also has many solutions) (Fig. 3.45). From the intersection points of the reference circles C 2 And O 3 draw straight lines, for example, through centers ABOUT And O 1 until they intersect with the reference circles at the junction points C, C 1 D And D 1, and radii R2, equal to the diameter of the reference circle - the conjugation arc.

Figure 3.45 Figure 3.46

Constructing an oval along two specified axes AB and CD(Fig. 3.46). Below is one of many possible solutions. A segment is plotted on the vertical axis OE, equal to half the major axis AB. From the point WITH how to draw an arc with a radius from the center SE to the intersection with the line segment AC at the point E 1. Towards the middle of the segment AE 1 restore the perpendicular and mark the points of its intersection with the axes of the oval O 1 And 0 2 . Build points O 3 And 0 4 , symmetrical to the points O 1 And 0 2 relative to the axes CD And AB. Points O 1 And 0 3 will be the centers of reference circles of radius R1, equal to the segment About 1 A, and the points O2 And 0 4 - centers of conjugation arcs of radius R2, equal to the segment O 2 C. Straight lines connecting centers O 1 And 0 3 With O2 And 0 4 At the intersection with the oval, the connecting points will be determined.

In AutoCAD, an oval is constructed using two reference circles of the same radius, which:

1. have a point of contact;

2. intersect;

3. do not intersect.

Let's consider the first case. A segment OO 1 =2R is constructed, parallel to the X axis; at its ends (points O and O 1) the centers of two supporting circles of radius R and the centers of two auxiliary circles of radius R 1 =2R are placed. From the intersection points of the auxiliary circles O 2 and O 3, arcs CD and C 1 D 1 are built, respectively. The auxiliary circles are removed, then the inner parts of the support circles are cut off relative to the arcs CD and C 1 D 1. In Figure ъъ the resulting oval is highlighted with a thick line.